2010年密云区中考一模数学答案

2010年密云县初中毕业考试

数学试卷答案参考及评分标准

阅卷须知：

1．为便于阅卷，本试卷答案中有关解答题的推导步骤写得较为详细，阅卷时，只要考

生将主要过程正确写出即可．

2．若考生的解法与给出的解法不同，正确者可参照评分参考给分． 3．评分参考中所注分数，表示考生正确做到这一步应得的累加分数．

一、选择题（本题共32分，每小题4分） 题 号 答 案 1 A 2 D 3 D 4 C 5 C 6 B 7 C 8 A

二、填空题（本题共16分，每小题4分） 题 号 9 10 11 12 答 案 x?1 a(a?b)(a?b) 4 2π 三、解答题（本题共35分，每小题5分） 13．（本小题满分5分）

?1?解：8?2sin45??(2?π)0???

?3??22?2?2?1?3 ··························································································· 4分 2?1?2?2． ·············································································································· 5分

14．（本小题满分5分）

解：去括号，得5x?12≤8x?6． ··············································································· 1分

移项，得5x?8x≤?6?12． ················································································ 2分 合并，得?3x≤6． ······························································································· 3分 系数化为1，得x≥?2． ······················································································· 4分 不等式的解集在数轴上表示如图：

? 0 2 3 3 ?2?1 1 ················································································································································ 5分 ·

1

15．（本小题满分5分）

x?1x2解：原式? ······················································································ 3分 ?x(x?1)(x?1) ?x． ········································································································ 5分 x?116．（本小题满分5分）

证明：在正方形ABCD中，

知AB=AD=DC=BC，∠B=∠D=90O．-------------------------------------------------2分 ∵ AE=AF， ∴ AB-AE=AD-AF．

即 BE=DF． ·········································································································· 3分 在△BCE和△DCF中，

?BE?DF,???B??D, ?BC?DC.?∴ △BCE≌△DCF． ···································································································· 4分 ∴ CE＝CF． ················································································································ 5分 17．（本小题满分5分）

1)， 解：∵ 一次函数y?kx?3的图象经过点M(?2，∴ ?2k?3?1． ····································································································· 1分 解得 k??2． ········································································································ 2分 ∴ 此一次函数的解析式为y??2x?3． 3分 令y?0，可得x??3． 2∴ 一次函数的图象与x轴的交点坐标为??，··········································· 4分 0?． ·令x?0，可得y??3．

?3?2???3)． ·∴ 一次函数的图象与y轴的交点坐标为(0，············································· 5分

18．（本小题满分5分）

解：如图，∵ AC平分∠BAD，

∴ 把△ADC沿AC翻折得△AEC，

2

∴ AE=AD=9，CE=CD=10=BC．------------------------------------------------------2分 作CF⊥AB于点F．∴ EF=FB=

11BE=（AB-AE）=6．------------------------3分 22在Rt△BFC（或Rt△EFC）中，由勾股定理得 CF=8．----------------------------4分 在Rt△AFC中，由勾股定理得 AC=17．

∴ AC的长为17． -------------------------------------------------------------------------5分

19． （本小题满分5分）

（1）证明：如图，连结OD，则 OD?OB．

∴ ?CBA??ODB．

∵ AC=BC, ∴ ?CBA??A． ∴ ?ODB??A．

∵ OD∥AC，∴ ?ODE??CFE．

? ∵ DF?AC于F，∴ ?CFE?90．

?∴?ODE?90．∴ OD?EF．

∴ EF是⊙O的切线． ------------------------------------------------------------3分

( 2 ) 连结BG，∵BC是直径, ∴∠BGC=90=∠CFE． ∴ BG∥EF．∴ ?GBC??E．

设 CG?x，则 AG?AC?CG?6?x．

在Rt△BGA中,BG?AB?AG?8?(6?x)．

在Rt△BGC中, BG?BC?CG?6?x．

2222∴ 8?(6?x)?6?x．解得 x?22222?2222222．即 CG?． 33在Rt△BGC中,sin?GBC?∴ sin∠E?GC1? ． BC91． --------------------------------------------- --------------------------------5分 9四、解答题（本题共11分，第20题5分，第21题6分） 20．（本小题满分5分）

解：设商场第一次购进x套运动服，

由题意得：

6800032000??10． ····································································· 3分 2xx解这个方程，得x?200． 经检验，x?200是所列方程的根．

2x?x?2?200?200?600．

答：商场两次共购进这种运动服600套． ····································································· 5分

3

21．（本小题满分6分）

解：（1）甲种电子钟走时误差的平均数是：

1(1?3?4?4?2?2?2?1?1?2)?0； 10乙种电子钟走时误差的平均数是：

1(4?3?1?2?2?1?2?2?2?1)?0． 10∴ 两种电子钟走时误差的平均数都是0秒． ---------------------------------2分 （2）S甲?11[(1?0)2?(?3?0)2???(2?0)2]??60?6(s2)； 1010112S乙?[(4?0)2?(?3?0)2???(1?0)2]??6?4.8(s2)．

10102∴ 甲乙两种电子钟走时误差的方差分别是6s2和4．8s2．---------------------------4分 （3）我会用乙种电子钟，因为平均水平相同，且甲的方差比乙的大，说明乙的稳定

性更好，故乙种电子钟的质量更优．

五、解答题（本题共4分） 22．（本小题满分4分）

解：（1）同意．如图，设AD与EF交于点M，

由折叠知，∠BAD=∠CAD，

∠AME=∠AMF=90O． ------------------------------1分

∴ 根据三角形内角和定理得

∠AEF=∠AFE． ------------------------------------2分

∴ △AEF是等腰三角形． ················································································ 3分

（2）图⑤中??的大小是22．5o． ·········································································· 4分

六、解答题（本题共22分，第23题7分，第24题7分，第25题8分） 23．（本小题满分7分）

解：（1）将A?3，2?分别代入y?-----------------------------------------6分

k，y?ax中， xk，3a?2， 32 ∴ k?6，a?．

3得2? ∴ 反比例函数的表达式为：y? 正比例函数的表达式为y?6； x

2x． ····························································· 2分 3 （2）观察图象得，在第一象限内，当0?x?3时，

反比例函数的值大于正比例函数的值．--------------------------------------------4分

4

（3）BM?DM．

理由：∵ S△OMB?S△OAC?1?k?3， 2 ∴ S矩形OBDC?S四边形OADM?S△OMB?S△OAC?6?3?3?12．

OB?12． 即 OC? ∵ OC?3，

∴ OB?4． 即 n?4．

63?． n2333∴ MB?，MD?3??．

222∴ m?∴MB?MD． ··························································································· 7分

24．（本小题满分7分）

解：（1）A（0，2）， B（?3，1）． ············································································· 2分

121x?x?2； ······································································ 3分 22117?）顶点为（?，． ················································································ 4分 28（3）如图，过点B?作B?M⊥y轴于点M，过点B作BN⊥y轴于点N，过点C?作

（2）解析式为y?C?P⊥y 轴于点P．

在Rt△AB′M与Rt△BAN中，

∵ AB=AB′， ∠AB′M=∠BAN=90°-∠B′AM， ∴ Rt△AB′M≌Rt△BAN．

∴ B′M=AN=1，AM=BN=3， ∴ B′（1，?1）． 同理△AC′P≌△CAO，C′P=OA=2，AP=OC=1， 可得点C′（2，1）； 将点B′、C′的坐标代入y?121x?x?2， 22可知点B′、C′在抛物线上． ········································································· 7分 （事实上，点P与点N重合）

25．（本小题满分8分）

解：（1）如图①，过D作DG∥AB交BC于G点，则四边形ADGB是平行四边形．

∵ MN∥AB，∴ MN∥DG． ∴ BG?AD?3． ∴ GC?10?3?7．

5

本文来源：https://www.bwwdw.com/article/eaxd.html